Area of a triangle inside of other triangle

Question

$BK$ and $CM$ are angle bisectors of angles $B$ and $C$. Let $L$ be the intersection point. If $AB=39,BC=42$ and $CA=45,$ find the area of $\triangle CKL$.

a) $756$

b) $252$

c) $126$

d) $140$

We can find the area of triangle $ABC$ by Heron's formula: $$S=\sqrt{63\cdot 21\cdot 18\cdot 24}=756.$$ I am stuck here.

Answer 1

Here are two observations that can get you to the answer quickly -

i) As $L$ is the incenter of the triangle, altitude from $L$ to all sides of the triangle is the same and that is inradius of the triangle $( = r)$.

ii) $A = r \cdot s$ where $A$ is the area of the triangle, $r$ is the inradius of the triangle and $s$ is sub-perimeter.

You have already found that $s = 63$ and you used that to find $A = 756$.

So area of $\triangle CLK = \frac{1}{2}r \cdot CK = \displaystyle \frac{756}{2 \cdot 63} \times \frac{42 \cdot 45}{42+39} = 140$

Answer 2

Use angle bisector theorem to find ratio between area of ABC ($S$) and area of CLK ($S_1$). To be specific:

1. BK is angle bisector of angles B, so $\frac{AK}{KC}=\frac{AB}{BC}=\frac{13}{14}$
2. Because of $AK+KC=AC=45 \rightarrow KC=\frac{14}{27}.AC=\frac{70}{3}$
3. CL is angle bisector of angles C, so $\frac{LK}{LB}=\frac{CK}{BC}=\frac{5}{9}$
4. $\frac{S_1}{S}=\frac{KC}{AC}.\frac{KL}{BK}=\frac{14}{27}.\frac{5}{14}=\frac{5}{27}$

Therefore, $S_1=140$